5.2.3 - Redox and Electrode Potentials Flashcards
How would you go about balancing a half equation and use the example : Fe+2 to Fe+3
How would you go about balancing a half equation and use the example : MnO-4 to Mn+2 (- charge and 4 at the bottom)
How would i combine 2 half equations
balance the electrons then cancel them out and then write the equation out.
Write down the redox titrations between iron and acidified potassium manganate
Oxidation half-reaction (Iron(II) to Iron(III)):
Fe+2 →Fe+3 +e-
Reduction half-reaction (Manganate to Manganese(II)):
Reduction half-reaction (Manganate to Manganese(II)):
MnO4− + 8H++ 5e- →Mn+2 +4H2O
Overall Balanced Equation:
MnO4- + 5Fe2+ + 8H+→Mn+2 + 5Fe+3 + 4H2 O
Things you should know about the redox titration between Iron and acidified potassium manganate.
- Potassium as a spectector ions : K+ ions are only there to balance the MnO4- ions but they don’t react with the Fe2+ ion so aren’t inclued in the ionic eqution
- Colour change : If the Potassium permanganate is in the burette then we will see a colour change from colourless to pink as there will be excess manganate when reacting with Fe but if the magnate is in the conical flask the there will be a colour change from Purple to Colourless.
- We use dilute sulfuric acid this is because it supplies the reaction with sufficient H+ ions.
- We don’t use Con HCL because Cl- is more electronegative than Fe2+ so MnO4- ions would react with the Cl- ions to produce Cl2 which is toxic and we wont have any manganate to react with Fe.
- We don’t use nitric acid as the NO3- ions react with Fe2+ ions to form a more stable product which means that the magante ions don’t react with the Fe2+ ions which are used up.
Write down the 2 redox equation for the thiosulfate titration.
Explain how you would carry out the redox titration for I2/S2O32-
- Conical flask: React your unknown solution containing IO3- ions, Your solution containing **I- ions and your acid together.
- Reaction : The IO3- and **I- will react together to form I2 which will display a brown colour.
- Titration : Then titrate the Sodium thiosulfate (that’s in the burette** against the Iodine in the conical flask.
-
Reaction : The reaction between the sodium thiosulfate and the Iodine will produce iodine ions again.
Colour change : We would observe a colour change from Brown to straw yellow. - Starch indicator : To ensure that all the iodine has reacted we add a starch indicator which will turn the solution blue. We then continue adding the sodium thiosulfate until all the blue turns to colourless as this would indicate that all the iodine has reacted.
What electrode do we use for electrochemical cells
Metal or platinum since platinum is inert.
Draw and set up an electrochemical cell
- The two half cells are joined together to give a complete circuit:
- the two metals are joined with a wire (electrons flow through the wire)
- the two solutions are joined with a salt bridge (ions flow through the salt bridge)
- a voltmeter is often included in the circuit to allow the potential difference (emf) or (Ecell) to be measured
What does the E cell value tell us ?
How easily a half cell gives up electrons
How do we know which metal will be reduce and oxidises.
Electrons will always move to a leasts reactive metal.
- Therefore the more reactive metal will loose electrons(Oxidised).
- The electrons will then move through the wires and move into the less reactive metal solution where the more reactive metal will gain electrons (reduced).
- Also if the Electrode potential value is given then remember NO PRoblem.
- Most negative half cell = Oxidised (flip the equation that is given).
- Positive half cell = reduction
What does the salt bride do in the electrochemical cell and what does it contain.
- The salt bridge is used to connect up the circuit. The free moving ions conduct the charge.
- A salt bridge is usually made from a piece of filter paper soaked in potassium nitrate (KNO3)
Standard conditions for electrochemical cells
Explain the relation between the anode and cathode and the redox reaction.
- At the Anode - Oxidation takes place.
- At the Cathode - Reduction takes place.
How do we calculate standard emf or Ecell of that’s what you want to call it.
Ecell∘ =Ecathode∘ −Eanode∘
(In other words : Reduction - Oxidation)
How do we measure the Ecell of a half cell
- We measure them against a reference half cell called standard hydrogen electrode
Set up a standard Hydrogen electrode and state the conditions
Note : If your using H2SO4 to obtain H+ ions then you use 0.5mols instead of 1mol beacuse there are 2 hydorgens instead of 1(H2SO4 is dicrotic) or you just use 1mol of HCL.
Also Hydrigen gas arrow must point in
What happens when there are two solutions in a half cell. - Draw how it would look like
- We use a platinum electrode.
How do we work out the strongest oxidising agent from a table
- The most positive equation and the one element on the left.
How do we work out the strongest reducing agent from a table
The most negative equation and the element on the left would be the strongest reducing agent.
Setting up cell notation
Salt bride = ll
What value does the Ecell have to be for the reaction to be feasible.
+
- The more positive the e.m.f the
more likely the reaction is to occur.
How does concerntration affect the Ecell
- Increasing con of one side would cause the equilibrium to shift to the other side.
- Increasing the concentration of O2 and decreasing the concentration of Fe(s) would cause Ecell to increase
How does temp affect the Ecell
- we would want to decrease temp as Most Ecells are exothermic.
- The rate of reaction is so slow that it appears there is no reaction (e.g rusting)
- If the reaction has a high activation energy it may stop the reaction from happening.
